Group Theory

Table of Contents

Group theory is an area of Algebra that rigorously studies symmetries, or more abstractly, automorphisms.

1. Overview

2. Summary of important results

2.1. Classification of finite simple groups

3. Definition and basic results

A group is a set equipped with the following data

  • a binary operation ;
  • a unary operation ;
  • an identity element ,

such that the following hold:

  1. Associativity: for all , ;
  2. Identity: for all , ;
  3. Inverses: for all , .

The order of a group , denoted , is the cardinality of the underlying set.

The order of an element is the smallest natural number such that:

Often a group is denoted just by its underlying carrier set . Where the group operation is unambiguous, we shall often leave it out and write for .

3.1. Examples

The automorphisms of an object form a group under composition.

The integers form a group under addition, with 0 as the identity.

For any , is a group.

The symmetric group on elements is a group.

The dihedral group is the group of symmetries of the -gon.

A group is called abelian if for all , .

3.2. Subgroups and Lagrange's Theorem

Like usual algebraic objects, we define a subgroup to be a subset which is closed under the same group operations as its main group.

A subgroup of a group is a non empty subset such that the following hold:

  1. For all , ;
  2. For all , .

We write to say that is a subgroup of .

It should be clear that these properties imply that .

The following lemma gives equivalent criterion for a subset being a subgroup.

Let . Then, if and only if for all .

Suppose that . Since , we have . Since , we have .

Now suppose that . Then and so .

The group of rotations of an -gon are a subgroup of .

The alternating group is a subgroup of .

Let and . The left coset of determine by is the set

Similarly, the right coset of determined by is

The number of left cosets of a subgroup is called the index of in and is denoted .

Let be a subgroup. Then the cosets partition the group .

Suppose that and , for some . Then some simple algebra shows that and , from which we deduce that and for some . Then any is equal to , so .

Similarly, we can show that , and so . Therefore or . It is also clear that every is in , and so cosets partition .

Lagrange's Theorem. Let be a subgroup of a finite group . Then .

This follows from cosets partitioning their group, along with the fact that each coset has the same size.

It is also clear that the function from is a bijection, and so any two left cosets of H have the same cardinality.

Suppose is a finite group and . Then, .

The subgroup has order and so by Lagrange's Theorem, must divide .

Suppose is a finite group. For all , .

Since the order of divides the , . So .

Let be a finite group. If is prime, then is cyclic.

If is prime, the only subgroups are and . Since is non-trivial for non-identity elements , it must be the whole group. Therefore, is equal to a cyclic subgroup and is cyclic itself.

Lagrange's Theorem can be used to prove Fermat's second most famous theorem.

If is a prime and , then .

This holds trivially for , so assume . Then, is in the multiplicative group of units , which has order . Since raising an element to the order of the group is the identity, , and so .

Wilson's Theorem also follows as an application of Lagrange's Theorem.

If is a prime number then the following hold:

  1. In , only and are their own inverses.
  2. .

We prove both parts.

  1. Clearly, 1 and are their own inverses. Take . Neither nor are multiples of and as is prime, is not a multiple of . Therefore, , and so proving that is not its own inverse.
  2. This follows by a simple counting argument. Every element can be paired up with its inverse, which is also in that range. Therefore, the product .

\end{enumerate}

3.3. Group homomorphisms

We now turn to the morphisms between groups.

Let be groups. A function such that , for all is a group homomorphism.

Let be a group homomorphism. Then the following hold:

  1. and .
  2. If is injective, then .
  1. We have the following equalities hold:

Therefore, we can apply to both sides of the equation and cancel to get . Recalling that inverses are unique, . Applying the similar proof by multiplying on the left, we deduce that .

  1. Suppose that . Then, , and so by injectivity of , . Suppose there exists such that . Then, , which contradicts being the order of . Hence is the order of .

Let be the cyclic group of order , which is defined as the set of rotations of the equilateral -gon. If is a rotation by radians, then . The function

is a group homomorphism.

Let be groups and a group homomorphism. We say that is a group isomorphism if there is a group homomorphism such that and are inverse functions.

If there exists an isomorphism , we say and are isomorphic as groups, and write .

If is a prime number, then all groups of order are isomorphic.

Let be groups of order . Then, every non-identity element has order , since by Lagrange's Theorem every non-trivial subgroup is equal to the whole group. Choose a non-identity element and . The function mapping is well defined and is a bijective group homomorphism. Hence, .

All cyclic groups of order are isomorphic.

This is the same proof as Lemma~\ref{lem:group-theory:groups-of-order-p-are-isomorphic}.

Given a group homomorphism, we can always extract a new group from it.

Let be a group homomorphism. The kernel of is the subset .

The kernel of a homomorphism (denoted ) is a subgroup of .

Clearly, for all we have that , and as , is a subgroup of .

A group homomorphism is injective if and only if .

Supposing that is injective, as , all elements of the kernel (which are of the form ) must be equal to .

Assuming the kernel is trivial, implies , and so . Therefore, and is injective.

Let be a group homomorphism. The image of is defined as , and is denoted .

Let be a group homomorphism. The is a subgroup of .

Let . Then , for some . As , is non-empty and by the Subgroup Criterion, is a subgroup of .

Let be a group. The set of all isomorphisms forms a group with composition and inverses, and is called the automorphisms group of , denoted .

For prime , .

Let be groups. The product group is the group of pairs of elements , with group operations defined pointwise.

We also say that is the external direct product.

Let and be subgroups of a group , such that , and that for all , . Then we say that is the \indx{internal direct product} of and if .

If is the internal direct product of then each element of can be written uniquely in the form for some and . This gives rise to an isomorphism .

Since , for every , there does exist a pair such that . If another pair with this property existed, then and , meaning that since the left hand side belongs to and the right to , both must be equal to the identity, proving uniqueness. The map defined by this is then a bijection. Routine computation shows this is a group homomorphism.

For coprime , . The element must have order , hence is isomorphic to the cyclic group of order . If there is a such that , then would need to share a factor of both and , which would contradict coprimality.

4. Isomorphism theorems for groups

Arguably, the most important type of subgroup is the normal subgroup.

A subgroup is said to be \indx{normal} if for all . We denote this by .

A subgroup is normal if and only if for all , , we have .

Suppose that is a normal subgroup and let and . Since if and only if and , we have , as required.

Now, assuming that for all , , we have that . However, as , we must have that and so , proving that is normal in .

  1. For any group , the centre is normal, since every commutes with all elements in .
  2. Any subgroup of an abelian group is normal.

As left and right cosets are the same for a normal subgroup , we typically just refer to them as ``cosets'', choosing to work with left cosets if necessary.

Let be a normal subgroup. The set of cosets is denoted and has a canonical group structure associated with it, defined by and . Groups of this form are called quotient groups.

This group structure is well defined for normal subgroups, but not always for arbitrary subgroups. To show that it is well defined, suppose and . We need to show that . This follows by normality of :

Inverses is well defined, also by normality, for if , then we have the following equations:

Hence, , as required.

Let be a group and a subgroup. Then, if and only if is the kernel of some group homomorphism from .

Suppose that . Then, there is a canonical function mapping . This function is a group homomorphism, as

It is clear that the kernel of this morphism is .

Now, supposing that , for some , Lemma~\ref{lem:group-theory:kernels-are-subgroups} shows that . If and , then

and so , which proves that by Lemma~\ref{lem:group-theory:normal-subgroup-criterion}.

4.1. First Isomorphism Theorem for groups

Quotient groups have the a universal property.

Let be a normal subgroup, and be a group homomorphism with . Then there is a unique group homomorphism that makes the following diagram commute:

The map is defined by . It is a well defined function as if , , and so giving the desired result that . It is a group homomorphism as is a group homomorphism. If there was another map that made this diagram commute, then , and so , proving that .

This universal property can be seen categorically as saying the map is a coequaliser in the category of groups of the diagram:

This result essentially gives us the First Isomorphism Theorem for groups.

Let be a group homomorphism. Then, is a normal subgroup of , is a subgroup of and there is an isomorphism

In particular, if is surjective, then .

The map is given by Theorem~\ref{thm:group-theory:universal-property-of-quotient-groups}. It is clear that the image of is , since , and if is surjective, it witnesses an isomorphism .

There is a surjective group homomorphism . The kernel of is the set , so .

4.2. The Second and Third Isomorphism Theorems and the Correspondence Theorem for groups

The second and third isomorphism theorems gives us information about relating the subgroups structure of to the subgroup structure of .

Let be a group and let . Let be the canonical morphism from Theorem~\ref{thm:group-theory:universal-property-of-quotient-groups}, and let .

  1. , with .
  2. if and only if .
  1. It's clear that the preimage is a subset of that contains . Let . We see that , which is in since it is a subgroup. Hence by Lemma~\ref{lem:group-theory:subgroup-criterion}, it is a subgroup of .
  2. Supposing that and let and . We see that , and as the preimage is normal in , . Therefore, , and so .

Now, assume that and let , . by normality of and . Therefore, , as required. This completes the proof.

Let and as above. If , then . That is, all subgroups of that contain are pulled back from subgroups of .

It is clear that, . % If , then and there must exists such that . % This means that and so . % Therefore, and so , as required.

These two propositions allow us to prove the following:

Let be a group with and the canonical morphism. The map is a bijection between subgroups of containing and subgroups of . Under this bijection, normal subgroups match with normal subgroups; further, if are subgroups of , then if and only if .

Injectivity of the function follows immediately from Proposition~\ref{prp:group-theory:normal-subgroups-pullback}. % Part 1 of Proposition~\ref{prp:group-theory:preimage-of-subgroups-of-G/N} shows that the function is surjective (since and ). % Part 2 of Proposition~\ref{prp:group-theory:normal-subgroups-pullback} shows us that normal subgroups match up with normal subgroups. % Now, supposing that , it is clear that . % Suppose that and let . % There must be such that , and so . % This means , for some . % Hence, and .

Let be a group, subgroups of with and . Then .

The index , and . We need to show that , as . But, is which is exactly the set of cosets of in and so has size .

Since is abelian, all of its subgroups are normal. % We can determine all of the subgroups of by finding all of the subgroups of that contain . % These are and . % Under the correspondence, they give the subgroups , , and . % There is a surjective group homomorphism mapping . % It is well defined as if then and it is easy to show that it is a group homomorphism. % Theorem~\ref{thm:group-theory:first-isomorphism-theorem} gives us that , and likewise, .

The Third Isomorphism Theorem tells us how to work with groups of the form .

\label{thm:group-theory:the-third-isomorphism-theorem} If , with , then

We have the following diagram:

Since , and Theorem~\ref{thm:group-theory:universal-property-of-quotient-groups} gives us the existence of . % Since is surjective, is also surjective. % We now show that , and use Theorem~\ref{thm:group-theory:universal-property-of-quotient-groups} to deduce the result. % Let . % Then, , so . % Let . % Then, , so which means that . % This shows that and so .

What can we say about subgroups of that don't contain the normal subgroup ?

Let be a normal subgroup of a group and a subgroup of . Then, the following hold:

  1. is a subgroup of ;
  2. ;
  3. ;
  4. There exists an isomorphism .

  1. It is clear that . Let and . Then, . Since, , , , for some , we can deduce the following:

and so .

  1. is normal in since all elements of are also elements of .
  2. Let . Then, for all , . Since and , , and so , which proves .
  3. By the Correspondence Theorem (Theorem~\ref{thm:group-theory:the-correspondence-theorem}) and , we have a group homomorphism . The kernel of this map is the set , and the image is (follows from and ). Hence, the First Isomorphism Theorem (Theorem~\ref{thm:group-theory:first-isomorphism-theorem}) gives an isomorphism

Let . Then, .

We take the function . The first isomorphism theorem gives

Applying the Second Isomorphism Theorem (Theorem~\ref{thm:group-theory:second-isomorphism-theorem}) to the middle term in the isomorphism chain, we get that

This gives an isomorphism

The left hand side has cardinality , while the right hand side's cardinality is . This means , as required.

5. Group presentations

Every group has a presentation which describes it wholly. It is formed by taking a free group and quotienting selected elements to the identity.

The free group on generators is the group whose elements are words in the symbols , subject to the group axioms and all logical consequences. The group operation is concatenation. It is written as .

Let . The group generated by subject to the relations is the group with generators , subject to the group axioms, the rules and all logical consequences. We write this group and call this a presentation of the group.

Let . This group must then satisfy and so this is a presentation the free group.

Let be a group and an arbitrary set. There is a bijection between the sets of group homomorphisms and functions .

6. Sylow theory

The Sylow Theorems act as a close converse to Lagrange's theorem. The converse to Lagrange's theorem does not hold in general.

The alternating group has order , but has no subgroup of order 6.

The best "partial" converse is we have is Cauchy's Theorem:

Cauchy's Theorem: If is a prime that divides the order of , then has a subgroup of order .

We need to introduce some tools before we can prove this result.

6.1. Group actions

Let be a group and a set. A (left) (or ) on is a function

such that the following hold:

  1. The identity acts trivially: , for all ;
  2. Actions are associative: .

A morphism between two group actions on sets and is a function such that .

A left group action is equivalently a functor . The functor category is isomorphic to the category of -sets.

Any -set gives rise to two important constructions.

The orbit of a group action is a subset of defined as:

The stabilizer of under a group action is the subgroup of

Let be a group that acts on a set . Then, for all , the stabilizer of the group action is a subgroup of .

Clearly . If , then

Therefore, and it is a subgroup of .

Let be a field and let be a positive integer. Defining and , acts on by matrix multiplication.

Let be a positive integer and let , the th symmetric group. Let . Then acts on by .

Let act on . Then,

  1. The action induces an equivalence relation on defined by: if and only if there exists with ;
  2. The equivalence classes of this relation are the orbits.
  3. The distinct orbits in form a partition of .

It is routine to check that that is an equivalence relation and that the orbits are the equivalence classes.

The fundamental theorem about group actions is the following:

Orbit-Stabilizer Theorem: Let be a finite group acting on a set and let . Then

This follows from~\nameref{thm:group-theory:Lagranges-theorem}. By Lemma~\ref{lem:group-theory:stabilizer-is-subgroup}, we just need to show that each of the stabilizer's cosets correspond to a unique element of , and visa versa. The maps we construct are and . These maps are well defined and inverse to each other, so there is a bijection between the cosets of the stabilizer and . Hence, .

We have enough to prove~\nameref{thm:group-theory:cauchys-theorem}.

Cauchy's Theorem. Let be a group with order , and let be a prime number that divides . Our goal is to show that contains an element of order , which generates a subgroup of order in . Define the set . The cyclic group of order , , acts on this group by . Our goal is to show that there must exist a non-identity element which is fixed by this group action.

By a simple counting argument, we see that , which is a multiple of . By Lemma~\ref{lem:group-theory:orbits-partition}, the orbits must partition , and by The~\nameref{thm:group-theory:orbit-stabilizer}, each orbit must have size or . Choosing a representative for each orbit , we have that One of these orbits is a singleton containing the element , so we can write

Since the left hand side is a multiple of , the right hand side must be also. As is not a multiple of , there must exist representatives that have orbits of size . Denote one such element by . We must also have that , and so for some non-identity element . Therefore, , and so . If the order of was less than , then would need to be a multiple of the order. As is prime, this is not the case, and so .

Any group can act on its underlying set in two different ways. The first is by multiplication, with , and the second is by conjugation: .

Let be a group and let . The centralizer is the Stabilizer where acts on itself by conjugation. It is precisely the set of elements in that commute with : .

Let be a group and let . The conjugacy class of is the orbit of where acts on itself by conjugation. We denote it by .

Let be a finite group. For any , we have

and this both and divide .

Immediate from the \nameref{thm:group-theory:orbit-stabilizer}.

Let be a finite group. Then, there exists elements such that

By Lemma~\ref{lem:group-theory:orbits-partition}, the orbits of the conjugacy group action partition . These orbits are the conjugacy classes, hence we get the above equation.

Let be a prime number. A -group is a group such that each element has order a power of .

If is finite, then is a -group if and only if is a power of .

If is a power of , by~\nameref{thm:group-theory:Lagranges-theorem}, for all , has order a power of and hence so does . Suppose for contradiction that is a prime not equal to that divides . Then~\nameref{thm:group-theory:cauchys-theorem} determines there must be a subgroup of order , and as is prime, must have an element of order , which is a contradiction.

Let be a non-trivial finite -group. Then the centre .

Choose such that . By Corollary~\ref{cor:group-theory:centralizer-conjugacy-classes-divide-G}, each term in the sum divides , and so for each , there exists such that .

Noticing that if and only if , we can split the sum from the class equation up as follows:

As if and only if , each must be a power of . Since , we must have and so is non-trivial.

If is a group with cyclic, then is abelian.

As is cyclic, there is some such that generates . Then, since the cosets partition , every element is equal to , for some . The product is then equal to for some . Each term in this product commutes with the other terms, since we can swap powers of and the centre of a group is exactly the elements that commute with every other element. Hence, , and is abelian.

If is a group with , then is abelian.

By Theorem~\ref{thm:group-theory:p-group-non-trivial-centre}, has a non-trivial centre, and so is equal to or . If it equals , then is abelian, so assume the order of the centre is . Since is normal in , we can quotient by it and get that has order and is therefore cyclic. By Lemma~\ref{lem:group-theory:cyclic-centre-implies-abelian}, must be abelian.

6.1.1. Polya counting

Before moving on to Sylow theory, it is worth looking at a beautiful application of group theory.

Let be a finite group acting on a finite set . For , define . Then, the number of orbit classes in is equal to .

Let . We see that for each , there are possible , and so . Also, for each , there are possible such that , so . By the \nameref{thm:group-theory:orbit-stabilizer} theorem, , and so . But the right hand side of this is exactly equal to the number of orbits classes of the group action.

How many essentially different ways are there of colouring the vertices of a regular 7-gon, with three colours? By ``essentially different'', we mean we quotient out by orbits of the canonical action of . Our goal then, is to count the number of orbits of acting on the set of all possible colourings of the 7-gon. There are total possible colourings.

  • The identity element fixes all coloured 7-gons, so ;
  • Any of the 6 non-trivial rotations only fix a 7-gon if every node has the same colour as every other node, of which there are three such 7-gons.
  • Any of the reflections (of which there are 7) only fix a 7-gon if each node's opposite in the line of symmetry is the same colour as itself (with one elment being its own opposite). This means that each reflection fixes exactly 7-gons.

The total then, by the \nameref{thm:group-theory:polya-enumeration-theorem} is

6.2. The Sylow Theorems

Let be a finite group and let be a prime number. A subgroup is a -subgroup of if it is a -group. We say it is a Sylow -subgroup of if its order is the highest power of that divides the order of . We say that is a Sylow subgroup of if its is a Sylow -subgroup for some prime .

If does not divide , then the trivial subgroup is the unique Sylow -subgroup of . When we want to exclude this case, we refer to .

For any group , let be the number of distinct Sylow -subgroups of .

It's not obvious that non-trivial Sylow -subgroups always exist. The first Sylow Theorem guarantees that they do. The second Sylow Theorem says that for a given all Sylow -subgroups of are conjugate, and the third gives information about the number of Sylow -subgroups of .

Sylow1: Let and be a prime that divides . Write , with . Then, there exists a subgroup of order ; that is, there is exists a Sylow -subgroup.

Define a set . This is a set of \textit{subsets} of . There is a group action we can put on , namely . To prove the First Sylow Theorem, we show that this group action has an orbit whose size is not divisible by . From this, we can then construct a subgroup of with order .

Since is formed from all subsets of size from a set of size , . We can write

Looking at the terms and for , we see that the highest power of dividing them both is the same - it is the highest power of that divides . We can then pair up terms on the numerator and denominator that are divisible by the same powers of . This results in the right hand side of the above equation not being divisible by , and so isn't either. As is a disjoint union of orbits, there must be at least one orbit whose cardinality isn't divisible by .

Choose an orbit whose cardinality is not divisible by and call it . The~\nameref{thm:group-theory:orbit-stabilizer} tells us that , which means that must be divisible by . Since , for any we have . This means . Therefore, is a subgroup with size .

Before proving~\ref{thm:group-theory:Sylow-2}, we prove the following lemma.

Let be a prime and a finite -group acting on a finite set . The number of fixed points in is congruent to .

Let be representatives for the orbits which partition . The singleton orbits are exactly the sets containing the fixed points of the group action. Define to be the subset of consisting of the fixed points, and remove and re-index our original representatives to be representatives of the remaining orbits: . As each is not a fixed point, the stabilizers are proper subgroups of and their indices must all be greater than 1, by~\nameref{thm:group-theory:Lagranges-theorem}. Each indice must divide the order of , and hence be a power of . The~\ref{thm:group-theory:orbit-stabilizer} then gives that divides for each . Since

and each of are divisible by , .

The proof of Theorem~\ref{thm:group-theory:p-group-non-trivial-centre} is really just an application of this proof.

We can now prove~\nameref{thm:group-theory:Sylow-2}.

Sylow 2: Let and be a prime that divides . Write , with . Suppose that is a Sylow -subgroup and is any -subgroup of . Then, there exists with . In particular, any two Sylow -subgroups of are conjugate in .

The -subgroup acts on the group of cosets by . Lemma~\ref{lem:group-theory:fixed-points-p-group-action} ensures that the number of fixed points of this action is congruent modulo to , by~\nameref{thm:group-theory:Lagranges-theorem}. Since can't divide as is a Sylow -subgroup, there must exist a fixed point, . Then, for all and so for all . Therefore, , and so .

Supposing that is a Sylow -subgroup, . Since , we must have and so is conjugate to .

A useful lemma this often used in conjunction with the Sylow Theorems is the following, which we can prove now with \nameref{thm:group-theory:Sylow-2}.

Let be a finite group and a prime number. Then has a Sylow -subgroup that is normal in if and only if .

Let be a Sylow -subgroup. By \nameref{thm:group-theory:Sylow-2}, all -subgroups are conjugate, so is the number of conjugates of and a subgroup is normal if and only if it has a unique conjugate. Hence, is normal if and only if .

The Third Sylow theorem tells us how to count Sylow -subgroups. This number is closely related to a new subgroup of .

Let be a group and . The normalizer of is

For any , is a subgroup of .

It is clear that . If , then we have and so .

It is clear that and that is the largest subgroup of with . The normalizer tells us how close a subgroup is to being normal, with if and only if .

Let be a finite group.

  1. For any subgroup , we have

  2. Let and let be a Sylow -subgroup of . Then .
  1. Let be the set of conjugates of . Then, acts on by conjugation with . By The \nameref{thm:group-theory:orbit-stabilizer} Theorem,

By definition, and . This gives

  1. This is immediate since the number of conjugates of is equal to , by the proof of Lemma~\ref{lem:group-theory:p-group-normal-iff-np=1}.

We can now prove \nameref{thm:group-theory:Sylow-3}.

Sylow 3: Let and be a prime that divides . Write , with . Then, and .

As , we have and so .

Let be the set of all Sylow -subgroups of and choose . Let the -group act on by conjugation. The number of fixed points for this action is congruent to , by Lemma~\ref{lem:group-theory:fixed-points-p-group-action}. We prove that is the unique fixed point of this action. It is clear that is a fixed point, since for all . If for all , then , and so and are Sylow -subgroups of . As , is the only Sylow -subgroup of by Lemma~\ref{lem:group-theory:p-group-normal-iff-np=1}. Then, we must have .

By Lemma~\ref{lem:group-theory:fixed-points-p-group-action}, , as required.

6.2.1. Applications

The Sylow theorems are very useful when trying to characterise finite groups. This section presents common uses of the theorems.

Consider , whose order is 6. It must have non-trivial Sylow -subgroups of order 2 and 3. There are 3 transpositions in , giving three Sylow 2-subgroups. Clearly, they are all conjugate. There is a unique subgroup of order 3, which is therefore normal.

Consider , the group of symmetries of the hexagon. It has 12 elements, so \nameref{thm:group-theory:Sylow-1} predicts the existence of a subgroup of order 3, and a subgroup of order 4. The subgroup of order 3 is generated by a rotation of degrees, while the subgroup of order 4 must by isomorphic to . For any reflection , we have , giving as a subgroup of order 4. There are three such subgroups of order four, each of them conjugate to the others.

Any group of order 30 has a nontrivial normal subgroup.

As , we have and , which means that there is either or . By Lemma~\ref{lem:group-theory:normal-subgroup-criterion}, we can assume that . The intersection of any two distinct Sylow 5-subgroups must be trivial, since the intersection of subgroups is a subgroup, and if the intersection contained an element of order 5, the subgroups would not be distinct. This means there are elements of order 5. Likewise, we have and and so either or . Without loss of generality, assuming that gives elements of order 3. This is means that our group of order 30 must have at least 44 elements in, which is plainly false. Therefore, either or , giving a non-trivial normal subgroup.

Let be a prime number and let for . Then must have a non-trivial normal subgroup.

Supposing that is not abelian, Theorem~\ref{thm:group-theory:p-group-non-trivial-centre} tells us that its centre is non-trivial. The centre is always a normal subgroup, so has a non-trivial normal subgroup. If is abelian, every subgroup is normal, including the subgroups of order that must exist by \nameref{thm:group-theory:cauchys-theorem}.

Let be a group with order for two distinct primes . Then, must have a non-trivial normal subgroup.

By \nameref{thm:group-theory:Sylow-3}, and . Either , or . If , there is a non-trivial normal subgroup by Lemma~\ref{lem:group-theory:p-group-normal-iff-np=1}, so assume that . This means there are elements of order . We also have and so assume that . This gives elements of order . In total we have elements. This is always greater than , either or is 1.

Let be a group with order for distinct primes and . Then, contains a non-trivial normal subgroup.

\nameref{thm:group-theory:Sylow-3} gives us that and . This means that is either 1 or . Assuming , there must be elements of order or . We also have that and , giving equal to either 1, or . If , there are elements of order . The sum is always greater than , so we must have , but this only adds more elements of order . Hence either or and there is a non-trivial normal subgroup.

7. Finitely generated abelian groups

This section will present results about characterising finitely generated abelian groups. The main results will have their proofs omitted, since they can be proved by more general results about -modules.

7.1. Finite abelian groups

Suppose that is a finite abelian group of order and that , for distinct primes . Let be the unique Sylow -subgroup of . Then That is, is isomorphic to the direct products of its Sylow subgroups.

Let be an abelian group with for some prime . Then is an internal direct product of cyclic subgroups of orders where and .

Let be a finite abelian group. Then is a direct product of cyclic groups of prime power order.

Let be non-zero coprime integers. Then .

Any finite abelian group of order can be written as a direct product of cyclic groups where for each and .

The \indx{exponent}, , of a finite group is the least common multiple of the orders of the elements of .

By \nameref{thm:group-theory:Lagranges-theorem}, .

If is a finite abelian group, then contains an element of order .

% TODO

If is a finite abelian group with then is cyclic.

Let be a finite subgroup of the multiplicative group of a field . Then is a cyclic group.

The subgroup must be abelian. Let and so for all , , so every element of is a solution to the polynomial . However, a polynomial of degree has at most roots in and so , which implies and so is cyclic.

The multiplicative group of a finite field is cyclic.

8. Symmetric and alternating groups

8.1. Symmetric groups

Symmetric groups are the group of permutations on objects. There are two different ways to represent permutations. A permutation can be represented as a array:

\begin{equation} \begin{bmatrix} 1 & 2 & \ldots & n \\ \sigma(1) & \sigma(2) & \ldots & \sigma(n) \end{bmatrix} \end{equation} 1

The other way is using \emph{cycle notation}, for example , which means the permutation sending . This is not unique. Cycles can be composed as expected. Two cycles are disjoint if no integer appears in both cycles. Two disjoint cycles commute, but this does not happen for arbitrary cycles.

Every permutation can be written as a product of disjoint cycles and the product is unique up to re-ordering the factors.

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A 2-cycle is called a transposition.

Every permutation can be written as the product of transpositions. This, is generated by transpositions.

Suppose that is a product of disjoint cycles of lengths , with . Then, the -tuple is called the \indx{cycle type} of .

Two permutations in are conjugate if and only if they have the same cycle type.

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8.2. Alternating groups

We can define a predicate on as follows: let be indeterminates and set Set . The symmetric group acts on by permuting the variables.

If fixes then is an \indx{even permutation}, while if then is an \indx{odd permutation}. The set of even permutations is denoted and is called the \indx{alternating group}.

The following hold:

  1. The product of two even permutations is even.
  2. The product of two odd permutations is even.
  3. The product of an even and odd permutation (in either order) is odd.
  4. A cycle length is even if is odd and is odd if is even.

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Let . Then, and hence has index 2. The size of is therefore .

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% TODO more results here

8.3. Simplicity of

A group is \indx{simple} if its only normal subgroups are and .

In this section, we prove that is simple when .

The alternating group is simple.

We show that there are 5 conjugacy classes in , containing 1, 15, 20, 12, 12 elements, respectively. If , then will be equal to a sum from the previous numbers\footnote{I think this follows from~\nameref{cor:group-theory:class-equation}}. This sum must then divide 60 (by~\nameref{thm:group-theory:Lagranges-theorem}) and the only possibilities for this are or , hence is simple.

For proving the result about all , we need a few lemmas.

If and are 3-cycles in , then and are conjugate in . That is, there exists with .

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If , then is generated by 3-cycles.

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Let . We say is \indx{fixed-point-free} if for all .

If and has the property that any non-identity is fixed-point-free, then .

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Let . Then is simple.

9. Jordan-H\"older

This section is about unique factorisation for groups.

9.1. Composition series

Let be a group. A \indx{composition series} for is a chain of subgroups

where and is simple for all .

We say that is the \indx{length} of the composition series and the simple groups are the \indx{composition factors}.

Here are some examples of composition series:

  1. .
  2. .

Let be a finite group. Then has a composition series. Moreover, any two composition series have the same composition length and they have the same composition factors up to isomorphism of groups and order of the factors.

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Let be a finite group. Then is isomorphic to one of:

  • Family 1: for prime;
  • Family 2: for with ;
  • 16 other infinite families;
  • 26 sporadic groups.

9.1.1. Existence

Suppose is a finite group with . Let be a composition series for be a composition series for , and be a composition series for . Then there is a composition series for of length whose composition factors are, in order,

As each is a subgroup of , the~\nameref{thm:group-theory:the-correspondence-theorem} gives us that is a subgroup of containing . Defining , we see that since , . the~\nameref{thm:group-theory:the-third-isomorphism-theorem} then gives that By assumption, is simple, and so is a composition series for .

If is a finite group, then has a composition series.

We proceed by induction. If is simple, then is a composition series. If is not simple, then contains a normal subgroup with . By Lemma~\ref{lem:group-theory:composing-composition-series}, the composition series for can be used to construct one for .

9.1.2. Uniqueness

The proof of uniqueness looks like a real pain. I will omit it for now.

Let be a finite group. Then any two composition series have the same length and the same composition factors up to isomorphism and the order in which they are listed.

More precisely, if and are two composition series for , then and there is a permutation of such that , for all .

10. Solvable groups

Solvable groups are groups that can be built up using compositions series, from abelian groups. They play an important role in Galois theory.

Let be a group. A \indx{subnormal series} for is a series of subgroups

ref:Group

A group is \indx{solvable} (or \indx{soluble}) provided that it has a subnormal series such that each factor is abelian.

  1. Any abelian group is solvable since is a suitable subnormal series.
  2. The group is not abelian, but it is solvable, as the subnormal series is suitable.
  3. The group is solvable.
  4. The group is not solvable as it is simple and not abelian.
  5. Any finite -group is solvable.

Solvability is related to composition series.

If is a finite abelian group of order , then the composition factors of are in some order.

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A finite group G is solvable if and only if all the composition factors of are cyclic.

If the composition factors are cyclic, they are abelian, and is solvable. Let be solvable and let be a subnormal series with abelian factors and assume (without loss of generality that ). By induction, the composition factors of are cyclic and is abelian. By Lemma~\ref{lem:group-theory:composition-factors-are-cyclic}, the composition factors are cyclic.

Let be a group and let . Then is solvable if and only if and are solvable.

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If is solvable and then is solvable.

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10.1. Derived subgroups and the derived series

Let be a group. The \indx{commutator} of two elements is the element , ans is often denoted by . The \indx{commutator subgroup} or \indx{derived subgroup} of a group is the subgroup generated by all possible commutators in . That is,

Let be a group and let be a normal subgroup of . Then is abelian if and only if . In particular, is abelian.

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Let be a group. Set and for each , set . The sequence is called the \indx{derived series} of .

A group is solvable if and only if there is an with .

% TODO

Let be a solvable group. Then for some . The least such is the \indx{derived length} of .

Let be a finite group with odd order. Then is solvable.

11. Questions and solutions

\label{sec:group-theory:questions}

11.1. Lagrange and group homomorphisms

\begin{questions} \begin{question} Let \(H\) and \(K\) be subgroups of a group \(G\). Show that \(H \cap K\) is a subgroup of \(G\). When is \(H \cup K\) a subgroup of \(G\)? \end{question} \begin{solution} It is clear that if \(h \in H \cap K\), then \(h^{-1} \in H\) and \(k^{-1} \in K\), hence \(h^{-1} \in H \cap K\). Therefore, for all \(h, k \in H \cap K\), \(hk^{-1} \in H \cap K\). \(H \cup K\) is a subgroup of \(G\) precisely when \(H \subseteq K\) or \(K \subseteq H\). For the sake of contradiction, assume \(H \not\subseteq K\) and \(K \not\subseteq H\). This means that there exists \(h \in H\) such that \(h \not\in K\) and \(k \in K\) such that \(k \not\in H\). Assuming that \(H \cup K\) is a subgroup of \(G\), \(hk \in H \cup K\) and so \(hk \in H\) or \(hk \in K\). This implies that either \(k \in H\) or \(h \in K\), which is a contradiction. The reverse implication holds clearly. \end{solution} \begin{question} Let \(\mu_{8}\) be the set of eighth roots of unity in \(\C^{*}\). What is the order of \(\mu_{8}\)? Find the elements \(g \in \mu_{8}\) such that \(\langle g \rangle = \mu_{8}\). \end{question} \begin{solution} The set of eighth roots of unity is \(\{z \in \C^{*} \mid z^{8} = 1\}\). There are exactly eight roots, by the fundamental theorem of algebra. Let \(\omega = e^{2i\pi/8}\). The generators of \(\mu_8\) are then \(\{\omega, \omega^{3}, \omega^{5}, \omega^{7}\}\). This is because these are the only elements of \(\mu_{8}\) whose exponents are coprime to \(8\). \end{solution} \begin{question} Let \(F\) be a finite field with \(q\) elements, and \(n \in \N\). Show that the order of \(GL(n, F)\) is \(\prod_{i = 0}^{n-1}(q^{n} - q^{i})\). \end{question} \begin{solution} Let \(A\) be an \(n \times n\) invertible matrix over \(F\). The first row can be anything except a zero row, so there are \(q^{n} - 1\) possibilities. For the next row, it can't be a scalar multiple of the first row, so there are \(q^{n} - q\) possibilities for the second row. The next row can't be a linear combination of the previous two rows and there are exactly \(q^{2}\) ways to do that, so there are \(q^{n} - q^{2}\) options. Generalising the pattern, we see that there are \[\prod_{i = 1}^{n-1} (q^{n} - q ^{i}),\] as required. \end{solution} \end{questions}

11.2. Isomorphism theorems

\begin{questions} \begin{question} The \indx{special orthogonal group} \(SO(n)\) is defined as the set of all \(2 \times 2\) real orthogonal matrices \(A\) with \(\det(A) = 1\). Show that every element of \(SO(2)\) has the form \[\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}\] for some \(\theta \in \R\). \end{question} \begin{solution} % TODO \end{solution} \begin{question} Show that \(SO(2)\) and \(\mu = \{z \in \C \mid |\| = 1 \}\). \end{question} \begin{solution} The isomorphism between the two groups is the function \[\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \mapsto e^{i (\theta \mod 2\pi)}\] \end{solution} We need to take theta \(\mod 2 \pi\) to make the homomorphism injective. \begin{question} Consider \(\mathbb{Z}\) and \(\R\) as groups under addiction. Prove that \(\R / \mathbb{Z} \cong \mu\). \end{question} \begin{solution} Take the function \(x \mapsto e^{ix2\pi}\) from \(\R \to \mu\). This is a surjective group homomorphism with kernel equal to \(\mathbb{Z}\). The~\nameref{thm:group-theory:first-isomorphism-theorem} gives us \(\R/\mathbb{Z} \cong \mu\). \end{solution} \begin{question} Find all subgroups of \(\mathbb{Z}, +\) that contain \(60\mathbb{Z}\) and find all subgroups of \(\mathbb{Z} / 60\mathbb{Z}\) and draw a diagram indicating the inclusions among the subgroups. \end{question} \begin{solution} We use the~\nameref{thm:group-theory:the-correspondence-theorem} to map between subgroups of \(\mathbb{Z}/60\mathbb{Z}\) and subgroups containing \(60\mathbb{Z}\). \begin{center} \begin{tikzcd} & 60\mathbb{Z} & & & & \mathbb{Z}/60\mathbb{Z} & & \\ 12\mathbb{Z} \arrow[ru, no head] & 20\mathbb{Z} \arrow[u, no head] & 30\mathbb{Z} \arrow[lu, no head] & & \mathbb{Z}/12\mathbb{Z} \arrow[ru, no head] & \mathbb{Z}/20\mathbb{Z} \arrow[u, no head] & \mathbb{Z}/30\mathbb{Z} \arrow[lu, no head] \arrow[ld, no head] & \\ 4\mathbb{Z} \arrow[u, no head] \arrow[ru, no head] & 6\mathbb{Z} \arrow[lu, no head] \arrow[ru, no head] & 10\mathbb{Z} \arrow[lu, no head] \arrow[u, no head] & 15\mathbb{Z} \arrow[lu, no head] & \mathbb{Z}/4\mathbb{Z} \arrow[u, no head] \arrow[ru, no head] & \mathbb{Z}/6\mathbb{Z} \arrow[lu, no head] & \mathbb{Z}/10\mathbb{Z} \arrow[u, no head] \arrow[lu, no head] & \mathbb{Z}/15\mathbb{Z} \arrow[lu, no head] \\ & 2\mathbb{Z} \arrow[lu, no head] \arrow[u, no head] \arrow[ru, no head] & 3\mathbb{Z} \arrow[lu, no head] \arrow[ru, no head] & 5\mathbb{Z} \arrow[lu, no head] \arrow[u, no head] & & \mathbb{Z}/2\mathbb{Z} \arrow[lu, no head] \arrow[u, no head] \arrow[ru, no head] & \mathbb{Z}/3\mathbb{Z} \arrow[u, no head] \arrow[ru, no head] \arrow[lu, no head] & \mathbb{Z}/5\mathbb{Z} \arrow[lu, no head] \arrow[u, no head] \\ & & \mathbb{Z} \arrow[lu, no head] \arrow[u, no head] \arrow[ru, no head] & & & & \{e\} \arrow[lu, no head] \arrow[u, no head] \arrow[ru, no head] & \end{tikzcd} \end{center} \end{solution} \begin{question} Let \(Z\) denote the centre of \(SL(n, K)\). Then the \indx{projective special linear group} \(PSL(n, K)\) is defined to be the quotient group \(SL(n, K) / Z\). Find the centre \(Z\) of \(SL(2, \mathbb{Z}_{3})\). What is the order of \(PSL(2, \mathbb{Z}_{3})\). \end{question} \begin{solution} Recall that the centre of a group is exactly the set of elements that commute with all other elements in the groups. The centre of \(SL(2, \mathbb{Z}_{3})\) consists of all matrices... % TODO:finish this solution \end{solution} \begin{question} Show that is \(G\) is a group, \(N\) is a subgroup of \(G\) contained in the centre of \(G\) and \(G/N\) is cyclic, then \(G\) is abelian. \end{question} \begin{solution} Let \(g, h \in G\). Then as cosets partition \(G\), \(g \in aN\) and \(h \in bN\), for some \(a,b \in G\). As \(G/N\) is cyclic, \(aN = k^{j}N\) and \(bN = k^{i}N\) for some \(k \in G\) and \(i,j \in \N\). Then, \begin{align*} gh &= k^{j}n_{1} k^{i}n_{2}, \\ &= k^{j} k^{i}n_{1} n_{2} & n_{1} \in Z(G), \\ &= k^{j} k^{i}n_{2} n_{1} & n_{2} \in Z(G), \\ &= k^{i} k^{j}n_{2} n_{1}, \\ &= k^{i} n_{2} k^{j} n_{1} & n_{2} \in Z(G) \\ &= hg. \end{align*} Therefore, \(G\) is abelian. \end{solution} \begin{question} Show that \(G \times \{e_{H}\}\) is a normal subgroup of \(G \times H\) that is isomorphic to \(G\) and that the quotient group \((G \times H)/G \times \{e_{H}\}\). \end{question} \begin{solution} To see that \(G \times \{e_{H}\}\) is normal, we see that it is the kernel of the map \((g,h) \mapsto h\) from \(G \times H \to H\). It is clear that it is isomorphic to \(G\) and that the map is surjective. By the~\nameref{thm:group-theory:first-isomorphism-theorem}, \((g \times H) / G \times \{e_{H}\} \cong H\). \end{solution} \end{questions}

11.3. Sylow theory

\begin{questions} \begin{question} Let \(p\) be a prime and \(G\) a \(p\)-group. Show that \(G\) contains an element of order \(p\). \end{question} \begin{solution} By Lemma~\ref{lem:group-theory:p-group-has-order-p^n}, \(|G| = p^{n}\) for some \(n \in \N\). Then, by~\nameref{thm:group-theory:cauchys-theorem}, \(G\) has a subgroup and hence an element of order \(p\). \end{solution} \begin{question} Let \(|G| = p^{n}\). Show that \(G\) contains a normal subgroup of order \(p^{i}\) for each \(i = 0, 1, \ldots, n\). \end{question} \begin{solution} This is true when \(n = 1\), so we proceed by strong induction. Consider \(G\) to have order \(p^{n}\). By theorem~\ref{thm:group-theory:p-group-non-trivial-centre}, \(G\) has a non-trivial centre, which must have order \(p^{m}\) for some \(0 < m < n\). By induction hypothesis, there are normal subgroups of orders \(p^{1}, \ldots, p^{m}\) in \(Z(G)\) which are therefore normal subgroups of \(G\), since all elements of \(Z(G)\) commute with all elements in \(G\). The quotient group \(G / Z(G)\) then has order \(p^{n - m}\), and so by induction hypothesis has normal subgroups of order \(p^{1}, \ldots, p^{n-m}\). By the~\nameref{thm:group-theory:the-correspondence-theorem}, these normal subgroups correspond to normal subgroups of \(G\) containing \(Z(G)\). Let \(H_{i}\) be the normal subgroup of \(G/Z(G)\) of order \(p^{i}\). Then by~\nameref{thm:group-theory:Lagranges-theorem}, \(p^{n-m} = [G/Z(G) : H_{i}] p^{i}\), and so \([G/Z(G) : H_{i}] = p^{n-m-i}\). By lemma \ref{lem:group-theory:index-of-quotients}, we have that \([G/Z(G) : H_{i}] = [G : \hat{H_{i}}] = p^{n-m-i}\), we have \(|\hat{H_{i}}| = p^{n - (n - m - i)} = p^{m + i}\) for \(i = 1, \ldots n-m\). This constructs all of the required normal subgroups of \(G\) groups. \end{solution} \end{questions}